Period POJ - 1961

Period

POJ - 1961

时限: 3000MS

内存: 30000KB

64位IO格式: %I64d & %I64u

提交 状态

已开启划词翻译

问题描述

For each prefix of a given string S with N characters (each character has an ASCII code between 97 and 126, inclusive), we want to know whether the prefix is a periodic string. That is, for each i (2 <= i <= N) we want to know the largest K > 1 (if there is one) such that the prefix of S with length i can be written as A K ,that is A concatenated K times, for some string A. Of course, we also want to know the period K.

输入

The input consists of several test cases. Each test case consists of two lines. The first one contains N (2 <= N <= 1 000 000) – the size of the string S.The second line contains the string S. The input file ends with a line, having the 
number zero on it.

输出

For each test case, output "Test case #" and the consecutive test case number on a single line; then, for each prefix with length i that has a period K > 1, output the prefix size i and the period K separated by a single space; the prefix sizes must be in increasing order. Print a blank line after each test case.

样例输入

3
aaa
12
aabaabaabaab
0

样例输出

Test case #1
2 2
3 3
 
Test case #2
2 2
6 2
9 3
12 4

来源

Southeastern Europe 2004

题意：一个字符串的前缀，可以由其最小的循环节循环k（k>1， 次得到。按升序输出这样的前缀的长度。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 
 using namespace std;
 
 #define maxn 1000008
 
 char s[maxn];
 int next[maxn];
 
 void getNext()
 {
     int j, k, i;
     i = strlen(s);
 
     j = ;
     k = -;
     next[] = -;
     while(j < i)
     {
         if(k == - || s[j] == s[k])
         {
             j++;
             k++;
             next[j] = k;
         }
         else
             k = next[k];
     }
 }
 
 int main()
 {
     int q, p = ;
     while(scanf("%d", &q), q)
     {
         scanf("%s", s);
 
         memset(next, , sizeof(next));
         getNext();
 
         printf("Test case #%d\n", p++);
 
         for(int i = ; i <= q; i++)
         {
             int k = next[i]; // 这个前缀的，前缀和后缀最长的相等长度
             if(i % (i - k) ==  && i / (i - k) != )  // 满足循环节……循环 输出
                 printf("%d %d\n", i, i / (i - k));
         }
         puts("");
     }
     return ;
 }